## Question 2: A circular drum of radius of 40 cm is initially rotating at 400 revolutions/min. It is brought to stop after making 50 revolution. What is the angular acceleration and the stopping time?

Explanation: We are given the initial angular velocity of the drum, 400 revolutions/min. Some opposite force is applied to stop the revolution of the drum. As the drum comes to stop, therefore, the final angular velocity is 0 rev/min. Similarly, the angular displacement before stopping is 50 revolution. We have to find the angular acceleration (off course it is negative angular acceleration) and the time in which drum stops (before taking the 50 revolutions.

Now for one revolution, the angular displacement is θ = 360° = 2π radian. Therefore, the angular displacement covered before stopping = 50 revolutions = 50 × 2π rad.

Given                    Radius of the drum = 40 cm = 0.4 m

Initial angular velocity, ωi = (400 rev × 2 × 3.14)/min = 2512 rev/60 s = 41.86 rev/s.

Final angular velocity, ωf = 0 rev/s

Angular displacement, θ = 50 rev = 50 × 2π rad

Angular acceleration, α

Formulae

For angular acceleration, we apply

2αθ = ωf2 – ωi2   …    (1)

For finding time, we apply,

ωf = ωi + αt          …    (2)

Note: All equations of motion in linear motion have their corresponding equations in circular motion. For, 2as = vf2 – vi2, the corresponding equation in rotatory motion is
2αθ = ωf2 – ωi2.
For vf = vi + at, the corresponding equation in circular motion is
ωf = ωi + αt

Solution

Put values in formula (1),

2× α(50 × 2 × 3.14) = 0 – (41.86)2 ⇒ 628α = – 1752.8

To find time, apply formula (2)

0 = 41.86 – 2.79t ⇒ -2.79t = -41.86

⇒ t = 41.86/2.79 ⇒ t = 14.99 s ≅ 15 s (Answer)